R - Race to 1 （期望）

Dilu have learned a new thing about integers, which is - any positive integer greater than 1 can be divided by at least one prime number less than or equal to that number. So, he is now playing with this property. He selects a number N. And he calls this D.
In each turn he randomly chooses a prime number less than or equal to D. If D is divisible by the prime number then he divides D by the prime number to obtain new D. Otherwise he keeps the old D. He repeats this procedure until D becomes 1. What is the expected number of moves required for N to become 1.
[We say that an integer is said to be prime if its divisible by exactly two different integers. So, 1 is not a prime, by definition. List of first few primes are 2, 3, 5, 7, 11, ]

 
Input
Input will start with an integer T (T <= 1000), which indicates the number of test cases. Each of the next T lines will contain one integer N (1 <= N <= 1000000).

 
Output
For each test case output a single line giving the case number followed by the expected number of turn required. Errors up to 1e-6 will be accepted.
 
Sample Input                             
3
1
3
13

Output for Sample Input
Case 1: 0.0000000000
Case 2: 2.0000000000
Case 3: 6.0000000000 
題解：設數x的期望爲dp[x]，1到x的素因子個數爲m,總素數爲n個。

則dp[x]=1+dp[x]*(1-m/n)+dp[x/y]/n;     (其中y爲1到x的素因子）（其實這個公式就是分兩種情況：1：該素數不是x的因子，概率爲（1-m/n）期望仍爲dp[x]. 2：該素數是x的因子，概率每個爲1/n，對應m個並由期望的線性性質得到：dp[x/y]/n，最後的1爲一次操作~

化簡得dp[x]=(n+dp[x/y])/m;

由於x必由小與x的數遞推而來，故可用遞歸來做。（打好素數表方便計算）

 

#include<bits/stdc++.h>
using namespace std;
const int maxn=1000010;
int vis[maxn],prime[maxn],tot;
double dp[maxn];//存儲答案
void GetPrime()//打好素數表
{
    memset(vis,0,sizeof(vis));
    for(int i=2;i<=maxn;i++)
    {
        if(!vis[i])
            prime[++tot]=i;
        for(int j=1;(j<=tot)&&(i*prime[j]<maxn);j++)
        {
            vis[i*prime[j]]=1;
            if(i%prime[j]==0)
                break;
        }
    }
}
double DFS(int x)
{
    if(x==1)//遞歸終止條件
        return 0;
    if(vis[x])//優化處理條件
        return dp[x];
    vis[x]=1;//每算出來一個標記一個
    int n=0,m=0;//1到x中總素數個數，素因子個數
    double ans=0;//答案
    for(int i=1;i<=tot&&prime[i]<=x;i++)
    {
        n++;
        if(x%prime[i]==0)
        {
            m++;
            ans+=DFS(x/prime[i]);//分量
        }
    }
    ans+=n;
    ans/=m;
    return dp[x]=ans;//記錄答案
}
int main()
{
    GetPrime();
    memset(vis,0,sizeof(vis));//打素數表的時候用到了vis，這時要重新清零
    int t,k=1;
    cin>>t;
    while(t--)
    {
        int x;
        cin>>x;
        cout<<"Case "<<k++<<": ";
        cout<<fixed<<setprecision(9)<<DFS(x)<<endl;
    }
    return 0;
}